 So we've seen that the mean free path, the average distance a molecule travels in the gas phase between collisions is given by this expression here. So if we know the temperature and the pressure and the size of the molecule, then we can calculate how far it travels through space before it collides with another molecule. But that raises the also interesting question, how much time passes before it collides with another molecule? How frequently in time do these molecules experience collisions? But we can answer that relatively easily. We know, first of all, we know velocity is distance traveled per unit time. We know the distance traveled by a molecule on average between collisions. We also know the average velocity of molecules in the gas phase. The average velocity and the mean free path are related to each other via the time between collisions. So I've written the time here as tau. Tau is the average time between collisions in the gas phase, so we could call that the collision time. So if I just rearrange that expression to get tau is equal to mean free path divided by the average velocity, and we know what both of those are. The mean free path is kT over square root of 2 and pi and d squared pressure in the denominator. The average velocity, average speed, I should say, remember is square root of 8 over pi kT over m. So this is in the denominator, so I'll write that upside down, square root of 8 under pi, and then the kT is in the denominator, the m is in the numerator, because it's upside down. So this is 1 over the average speed. Now if we just simplify that expression a little bit, let's see, I've got a square root of 2 and a square root of 8 in the denominator, so square root of 16 is 4 in the denominator. I've got a d squared and a p in the denominator, and then I think everything else will be inside a square root, because I have kT divided by square root of kT, so I end up with a kT in the numerator of the square root. I've got a square root of m in the numerator, and I have pi in the denominator, square root of pi in the numerator, so that leaves me with a pi in the denominator of that square root. So if I haven't missed any terms, that would be the average time between collisions. If we want to know that, we can work a quick example. So if we have nitrogen at 298 Kelvin and 1 atmosphere, and as we've seen previously the diameter of a nitrogen molecule when treated as, using the kinetic theory of gases, that diameter is about 3.6 angstrom, so if we wanted to know what's the average time between collisions for nitrogen molecule and the gas phase, we've got two choices. We could either plug into this expression, I'll do that once just to show how the units work out. So 3.6 angstroms for the diameter, that's 3.6 times 10 to the minus 10 meters, that quantity is squared. The pressure, 1 atmosphere, I'm going to want that in SI units, so I'll write that as 101,325 Pascals, or Joules per cubic meter, that's 1 atmosphere in units of Pascals. So that's my 1 over 4 D squared P inside the square root. As usual, when I have a mass and these expressions, I need the mass of one molecule. We certainly know mass of a nitrogen molecule in grams per mole, to make sure the units work out, we're going to need to convert grams to kilograms, and to get rid of the moles in the denominator here, one mole is 6 times 10 to the 23rd molecules. So that leaves me, double check the cancellations, but moles have canceled, grams on top cancel grams on the bottom, so I'm left with just kilograms, that's the mass of a molecule in kilograms. I can also multiply by K and T, Planck's constant temperature, that's all inside this great big square root. On the top of the fraction on the bottom, I've got a pi, so all I need down there is pi. So everything there is in SI units, Kelvin are going to cancel Kelvin, Joules is a kilogram meter squared per second squared, if I add another kilogram to give me kilogram squared, meter squared per second squared, then when I take the square root, that's going to have units of just one kilogram and one meter over one second. Down here I've got, let's see, meters squared, cancels two of these meters in the meter cube, so I end up with the one factor of meter in the denominator, the denominator moves it up to the top, and then the joule is on the bottom. That joule is a kilogram meter squared per second squared, so this joule cancels this kilogram meter squared per second, let's see, doesn't cancel a second squared. So let's see, if I write another second over a second, there's the second squared that's going to cancel, and I'm left with units of just seconds. So the units are going to work out fine, this average collision time is going to be in units of seconds, and if we do all the math, what we'll end up with is a value of 1.5 times 10 to the minus 10 in units of seconds. So 1.5 times 10 to the minus 10 seconds, that would be 150 picoseconds. A picosecond is 10 to the minus 12 seconds, so 150 picoseconds is a very short amount of time, that's on average the amount of time spent between collisions in the gas phase for nitrogen molecule at room temperature and pressure. I mentioned there were two options for how to do this calculation. This is maybe the most straightforward approach using the equation that we've derived for the collision time. We could also go back, since we've previously calculated the mean free path and the average velocity for a nitrogen molecule, we could have done it that way as well. We could have said the collision time is equal to the mean free path divided by the average velocity. And if we recall what those values were when we saw them previously under these same conditions, that was 70 nanometers for the mean free path and 474 meters per second for the average velocity. And if we convert the units, divide this number by this number, it's no surprise that that'll also come out to exactly 150 picoseconds. If you happen to have already calculated the mean free path and the average velocity, this math is, the arithmetic anyway, is significantly easier than using this full expression because we've already done the work of all these unit conversions and square roots in order to calculate these two numbers. So that's the average time between collisions. Often more useful than the average time between collisions is knowing the frequency of collisions. How many times per unit time does a molecule experience a collision? So the collision frequency, we can define as one over the average time. So one over the collision time is the collision frequency, the number of times per second that a molecule experiences a collision. So if the collision time was this expression, the collision frequency is just one over that expression. So one over 4d squared p becomes 4d squared p. The square root turns upside down. So I've got square root of pi over mkt. And that would be the expression for the collision frequency to show you what that looks like for our standard example. Now at this point, nitrogen at 298 Kelvin, one atmosphere of pressure, where the diameter of that molecule is 3.6 angstroms. We could plug into this expression, or again, since we've already calculated the collision time, collision frequency is one over that. So the collision frequency, one over 150 picoseconds, one over 150 times to the minus 12 seconds. That works out to a large number with units of one over seconds. 6.7 times 10 to the ninth, so 6.7 billion one over seconds. So the collision frequency is 6.7 billion times per second. If you picture a nitrogen molecule bouncing around in the gas phase, traveling roughly 70 nanometers between collisions, moving at this velocity, it's going to experience a collision with another molecule about 6.7 billion times per second. So these expressions, particularly the collision frequency, now that we know how many times the molecule collides with other molecules in the gas phase under a certain set of conditions, that will open the door to being able to calculate things about chemical reactions. If a molecule is going to react with another molecule in the gas phase, it needs to first collide with that molecule. So now that we know how often that's happening, we can make some progress towards understanding how quickly the reaction would happen. We're actually going to put that off until later in the semester, and in the meantime, we'll continue talking about other properties of gases.